Question: $\sum\limits_{k=0}^{{19}}{{{6(1.5)^{k}}}} \approx $ Choose 1 answer: Choose 1 answer: (Choice A) A $7983.02$ (Choice B) B $ 13{,}301.03 $ (Choice C) C $26{,}590.05 $ (Choice D) D $39{,}891.08$
Explanation: What is the question asking for? The question is asking for the sum of the values of $6(1.5)^k$ from $k = 0$ to $k = 19$ : $6(1.5)^0 + 6(1.5)^1 +... +6(1.5)^{19} $ The series is geometric because the formula $6(1.5)^k$ is an exponential function of $k$. Formula for geometric series The sum $S_n$ of a finite geometric series is $S_n = \dfrac{a_1(1-r^n)}{1-r}$ where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. What do we need to use the formula? The number of terms $n$ is ${20}$ because there are ${20}$ numbers from $0$ to $19$. The first term $a_1$ is $6$ because $6(1.5)^0 = {6}$. The common ratio $r$ is ${1.5}$ because it is the base of the exponent in $6({1.5})^k$. Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac{a_1(1-r^n)}{1-r} \\\\ S_{{20}}&=\dfrac{{6}(1-\left({1.5}\right)^{{20}})}{1-\left({1.5}\right)} \\\\ S_{{20}}&=-12{(1-\left({1.5}\right)^{{20}})}\\\\ S_{{{20}}} &\approx 39{,}891.08 \end{aligned}$ The answer $ 39{,}891.08 $